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Brilliant’s Composite Mathematics Class 6 Solutions Chapter 1 Revision of Work Done in Earlier Classes
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 1, Revision of Work Done in Earlier Classes. Here students can easily find step by step solutions of all the problems for Revision of Work Done in Earlier Classes. Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Revision of Work Done in Earlier Classes Solution :
Question no – (1)
Solution :
(i) 19,24, 50, 601
= 10000000 + 90000000 + 30000000 + 400000 + 50000 + 600 + 1
Place value of 6 = 600
(ii) 53, 42, 01, 602
= 500000000 + 30000000 + 4000000 + 200000 + 1000 + 600 + 2
Place value of 6 = 600
(iii) 27, 56, 225
= 2000000 + 700000 + 50000 + 6000 + 200 + 20 + 5
Place value of 6 = 6000
(iv) 2,00, 35, 206
= 20000000 + 700000 + 50000 + 6000 + 200 + 20 + 5
Place value of 6 = 6
Question no – (2)
Solution :
(i) 5,00,00,000 + 60,00,000 + 3,00,000 + 2,000 + 800 + 6
In the short form
= 5, 63, 02, 806
(ii) 1,00,00,00,000 + 50,00,00,000 + 40,00,000 + 8,00,000 + 70,000 + 9,000 + 500 + 20 + 9
In the short form
= 1150, 48, 79, 529
(iii) 7,00,00,000 + 80,00,000 + 6,00,000 + 2,000 + 100 + 30 + 1.
In the short form
= 7, 86, 02, 131
Question no – (3)
Solution :
(i) 57,39,605; 43,98,056; 57,38,635; 67,83,505
Now in ascending order,
= 43, 90, 056, 5738635, 5739605, 6783505
(ii) 33,85,60,705; 38,85,60,705; 33.85,70,705; 33.85,70,805
Now in ascending order,
= 3388560705, 338570705, 338570805, 388560705
Question no – (4)
Solution :
(i) 68,93,950; 68,83,950; 6,89,39,510; 75,13,501
Now in descending order,
= 6893510, 7513501, 6893950, 6883950
(ii) 55,39,839; 55, 38, 753 ; 55, 49, 839 ; 55, 48, 639
Now in descending order,
= 5549839, 5548639, 5539839, 5538753
Question no – (5)
Solution :
In the question we get,
Population of the Union territories of Delhi
= 62,20,406 and
Population of the Union territories of Chandigarh
= 6,04,471
∴ Combined population,
= 6220406 + 604471
= 6824877
Hence, the combined population will be 6824877
Question no – (6)
Solution :
In the question we get,
Teachers working in primary schools = 15,30,145
Teachers in upper primary schools = 10,11,049
Teachers in secondary schools = 7,25,935
Teachers in higher secondary schools = 4,25,622
∴ Total number of teachers,
= 1530145 + 1000049 + 725935 + 425622
= 3692751
Thus, total number of school teachers will be 3692751
Question no – (7)
Solution :
Given in the question,
The sum of two numbers = 8,38,52,612.
one of the numbers = 5,63,45,218
Other number = ?
∴ The other number,
= 8,38,52,612 – 5,63,45,218
= 2,75,07,394
Thus, the other number will be 2,75,07,394
Question no – (8)
Solution :
According to the question,
Factory produced switches in 1996
= 38,56,921
Factory produced switches on 1998
= 8,96,57,321
∴ Increase in the production of switches,
= 89657321 – 3856921
= 8,58,00,400 switches
Hence, the increase in the production of switches will be 8,58,00,400 switches.
Question no – (9)
Solution :
In the given question,
In a school number of students = 6,83,52,561
Girls student = 88,56,981
Boy students = ?
∴ Number of boy students,
= 68352561 – 8856981
= 5,94,9558 boys
Therefore, the number of boy students will be 5,94,9558
Question no – (10)
Solution :
In the given question we get,
In an examination in year 1998, students passed = 8,56,34,517
got first division = 88,93.651
got second division = 3,98,56,512
got third division = ?
∴ Students got third division,
= 8, 56, 34717 – (889651 + 3,98,56,512)
= 85634571 – 48750163
= 36684354
Hence, in exam 36684354 students got third division.
Question no – (11)
Solution :
(i) 4715125 by 543
∴ Quotient = 8683,
∴ Remainder = 256
Now, Check :
(8683 × 543) + 256
= 4715125
(ii) 3876 by 228
∴ Quotient = 17,
∴ Remainder = 0
Now, check :
= (17 × 228) = 0
= 38 + 6
(iii) 434350 by 238
∴ Quotient = 1825,
∴ Remainder = 0
Now, check,
= (1825 × 238) + 0
= 434350
(iv) 87325562 by 3405
∴ Quotient = 25646
∴ remainder = 932
Now, check :
= (25646 × 3405) + 932
= 87325562
(v) 470988 by 378
Quotient = 1246,
Remainder = 0
Now, check :
(1256 × 378) + 0
= 470988
(vi) 34705826 by 429
∴ Quotient = 80899,
∴ Remainder = 115
Now, check :
= (80800 × 429) + 155
= 34705826
Question no – (12)
Solution :
(i) In the given question,
Divisor 375,
quotient = 8751,
remainder = 78
As we know that,
Dividend = (Quotient × Division) + remainder
∴ Dividend,
= (8751 × 375) + 78
= 3281703
Thus, the dividend will be 3281703
(ii) In the question we get,
Divisor = 4081,
Quotient = 7631,
Remainder = 371,
As we know that,
Dividend = (Quotient × Division) + remainder
∴ Dividend,
= (7631 × 4081) + 371
= 31142482
Hence, dividend will be 31142482
Question no – (13)
Solution :
First, July + August + September,
= 31 + 31 + 30
= 92 days
Now, Sell every day,
= 38104/92
= 4142 litre
Question no – (14)
Solution :
In the given question,
A factory produces screws in a year = 26763750
screws are packed in a carton = 375
∴ Screws produced in a year,
= 26763750/375
= 71370 cartons
Hence, 71370 required to pack the screws produced in a year.
Question no – (15)
Solution :
575 trucks can carry,
= 735856 kg 250 g …(Given)
= 735856 kg + 250/1000 kg
= 735856 + 0.25
= 735856.25 kg
∴ 1 track can be carried,
= 7358625/513 × 100
= 12975/100
= 1279.75 kg
= 1279 kg 750 gm
So, 1 truck will carry 1279 kg 750 gm.
Question no – (16)
Solution :
Given in the question,
10342208 were divided equally among 513 cancer patients
∴ Each patient got,
= 10342308/513
= 20160.25
Hence, each patient will got Rs 20160.25
Question no – (17)
Solution :
He spend money,
= (1358 × 2) + 23568 + (2 × 635870)
= 2716 + 23568 + 1271740
= 129802
∴ Money will be left in his a account,
= 1856500 – 1298024
= 558476
Question no – (18)
Solution :
In 7 days the engine run,
= 12 × 7
= 84 hour
∴ Litters of water will it pump in 7 days,
= (84500 × 84)
= 7098000 litres
Thus, 7098000 litres of water will it pump out in 7 days.
Question no – (19)
Solution :
(i) 3/4 × 5 1/2 + 7 2/3 + 1 1/4 × 1 3/5
∴ 3/4 × 5 1/2 + 7 2/3 + 1 1/4 × 1 3/5
= 3/4 × 11/2 + 23/3 + 5/4 × 8/5
= 33/8 + 23/3 + 2
= 99 + 184/24 + 12
= 283 + 48/24
= 13 19/24
(ii) 4 1/2 × 3 3/5 – 6/7 × 1 2/4 + 4/5
∴ 4 1/2 × 3 3/5 – 6/7 × 1 2/4 + 4/5
= 4 1/2 × 3 3/5 – 6/7 × 1 2/4 + 4/5
= 9/2 × 18/5 – 6/7 × 6/4 + 4/5
= 81/5 – 9/107 + 4/5
= 85/5 – 9/7
= 119 – 9/7
= 15 5/7
Question no – (20)
Solution :
Given, 2/3 × (4/5 = 1/10) = 2/3 × 4/5 = 2/3 × 1/10
∴ 2/3 × 4/5
= 2/3 × 1/10
= 2/3 × 4/5
= 2/3 × 1/10
Question no – (21)
Solution :
In the given question,
The cost of a chair is Rs 115 1/2 and that of table is Rs 284 1/2
Money required to buy 25 chairs and 25 tables = ?
To buy 25 chairs required money,
= 115 1/2 × 2/5
= 231 × 25/2
To buy 25 tables required money,
= 284 1/2 × 25
= 569 × 25/2
∴ Total money,
= 231 × 25/2 + 569 × 25/2
= 25/2 (231 + 569)
= 35/2 × 800
= 10000
Thus, to buy 25 chairs and 25 tables Rs 10000 will be required.
Question no – (22)
Solution :
In the given question,
The weight of a bag of rice is = 38 3/4 kg.
To put into the bag = 1937 1/2 kg rice
number of bags needed = ?
∴ Number of bags,
= 1937 1/2/38 3/4 – 3875/2 × 4/15
= 50 bags
So, the number of bags needed is 50.
Question no – (23)
Solution :
Given in the question, 107 m 40
= 107 + 40/100 m
= 107/10 m
Now in fraction,
= 21/1074/10
= 210/107
= 135/179
Question no – (24)
Solution :
In the given question,
= 49 7/8
= 399/8
Now in fraction,
= 226/399/8
= 266 × 8/399
= 5 1/3
Question no – (25)
Solution :
(i) Given, 0.005 by 0.25
= 0.005 × 100/0.25 × 1000
= 0.02
(ii) Given, 0.00876 by 0.219.
Now,
= 0.00876 × 1000/0.219 × 100000
= 4/100
= 0.004
Question no – (26)
Solution :
Given in the question,
6 kg cauliflower = 33.60
1 kg tomato = 11.50
∴ 2 kg potato,
= 2 × 7.10
= 14.20
∴ 6 kg onion,
= 6 × 5.50
= 33.00
Now, she get back,
= 100 – (33,60 + 10.50 + 14.20 + 33)
= 100 – 92.30
= 7.70 Rs.
Hence, she will get back 7.70 Rs
Question no – (27)
Solution :
In the question we get,
Rs 2625 equally among = 500 children
Each child get = ?
∴ Each children get,
= 2625/500
= 525/100
= 5.25
Thus, each child will get 5.25 Rs
Question no – (28)
Solution :
In the question,
Salary of 75 servants for one mouth is = Rs 9502.50
Salary of each servant per month,
Hence, Salary of each servant per month will be 126.70 Rs
Question no – (29)
Solution :
Given, (5/7 – 1/9 of 1 2/7) + 2 4/5 2 1/7 1 1/7 – 5/7 5 × 2
∴ (5/7 – 1/9 of 1 2/7) + 2 4/5 2 1/7 1 1/7 – 5/7 5× 2
= (5/7 – 1/9 of 1 2/7) + 2 4/5 × 2 1/7 × 1 1/7 – 5/7 ÷ 5 × 2
= (5/7 – 1/9 of 9/7) + 14/5 × 15/7 × 8/7 – 5/7 × 1/8 × 2
= 4/7 + 48/7 – 2/7
= 4 + 48 – 2/7
= 7 1/7
Question no – (30)
Solution :
(i) 2 + {6 – { 2 (3 ÷ 6 + 1/9 of 15)}]
Now,
= 2 + {6 – { 2 (3 ÷ 6 1/9 of 15)}]
= 2 + {6 – {2 (3 ÷ 6 + 1/9 × 3)}]
= 2 + {6 – {2 } 3 ÷ 9}]
= 2 + {6 – { 2 × 3/9}]
= 2 + 6 – 6/2
= 6 + 18 – 6/9
= 6 + 18 – 2/3
= 7 3/1
(ii) 2 – [6 ÷ 2 { 6 × 1/2 = (3 1/2 – 1 1/2)}]
Now,
= 2 – [6 ÷ 2 { 6 × 1/2 = (3 1/2 – 1 1/2)}]
= 2 [6 ÷ 2 ÷} 3 = (7/2 – 3/2)}}
= 2 [6/2 ÷ } 3 = 4/2}]
= 2 [6/2 ÷ } 2}]
= 2 [6/2 ÷ 20
= 2 × 3/2
= 3
Question no – (31)
Solution :
| Rounded off to tens | Rounded off to hundreds | Rounded off 10 thousands | |
| (a) 2795 | 2800 | 2800 | 3000 |
| (b) 1635 | 1640 | 1600 | 2000 |
| (c) 8935 | 8940 | 8900 | 9000 |
| (d) 33561 | 33560 | 33600 | 34000 |
Question no – (32)
Solution :
In the given question,
Meenakshi added = 1087, 1321, 1663, 780 and 6312
∴ 1087 + 1321 + 1663 + 780 + 6312
= 11163
Yes, the solution appear to be sensible.
Question no – (34)
Solution :
In the question, 863 by 13
Now, 863 by 13
∴ 66.385
Question no – (35)
Solution :
Let, he score in 3rd innings = x
= 80 + 50 + x + 79 + 50/5 = 60
= x + 263 = 300
= x = 37
Hence, he should score 37 runs in 3rd innings.
Question no – (36)
Solution :
In the given question,
The ages of 5 members in a family are = 58, 54, 20, 18, 15.
Average age after 12 years. = ?
After 12 years family members ages,
= 70, 66, 32, 30, 27
Now, Average ages,
70 + 66 + 32 + 30 + 27/5
= 225/5
= 45 years
Thus, their average age after 12 years will be 45 years.
Question no – (37)
Solution :
In the question we get,
For 2 hours at a speed of = 70 km/hr.
For the next hour = 80 km/hr and
For the last 3 hours = 60 km/hr
∴ Average speed of the train,
= 2 × 70 + 80 + 3 × 60/6
= 140 + 80 + 180/6
= 400/6
= 66.667 km/hr
Therefore, the average speed of the train will be 66.667 km/hr.
Question no – (38)
Solution :
From the question we get,
Ritika obtained in English = 80 marks
In Hindi = 75 marks
In Science = 88 marks
In Mathematics = 85 marks
∴ Average marks,
= 30 + 75 + 88 + 85/4
= 8 328/4
= 82
Hence, the average marks of Ritika will be 82.
Question no – (39)
Solution :
The price of onions on 5 days were,
Rs 3 per kg,
Rs 4 per kg,
Rs 4.50 per kg,
Rs 3.50 per kg
Rs 5 per kg.
∴ Average price of onion,
= 3 + 4 + 4.50 + 3.50 + 5/5
= 20/5
= 4 rupees.
Hence, average price of onion will be 4 rupees.
Question no – (40)
Solution :
90 km goes in = 3h
1 km goes in = 3 × 90h
∴ 60 km goes in
= 3 × 90/200
= 4.5 hr
Hence, it take 4.5 hr to reach Delhi.
Question no – (41)
Solution :
From the question we know,
An earthworm travels 27 decimeters in 9 minutes
Now,
(a) 1 min travel,
= 27/9
= 3 decimetres
(b) 10 min travel,
= 3 × 10
= 30 decimetres
(c) 15 min travel,
= 3 × 15
= 45 decimeters
Question no – (42)
Solution :
From the question we know,
A child runs at a speed of 10 m/sec.
he take to cover a distance of 1 km = ?
Now, 10 m goes = 1 sec
1 m goes = 1/10 11
1000 m goes,
= 100 × 1/100 see
= 100 sec
Thus, he take 100 sec to cover a distance of 1 km.
Question no – (43)
Solution :
As we know that,
Speed = distance/time
∴ Speed will be,
= 216 × 1000/4 × 3600
= 15 m/s
So, her speed will be 15 m/s.
Question no – (44)
Solution :
Time = 150/60 = 2.5 h
= 2 hours 30 min
Stopped for 45 min
Left time = 2 h 30 min + 45 min = 3 h 15 min
∴ He reach = 11 : 39 am + 3h 15 min
= 2 : 54 min
= 2 : 54 P.M
Hence, he will reach his destination at 2 : 54 p.m.
Question no – (47)
Solution :
(a) 12 1/52% of 320 kg
= 320 × 2/100
= 40 kg
Thus, the value will be 40 kg.
(b) 5% of 180
∴ 180 × 5/100
= 9
Hence, the value will be 9
(c) 25% of 160
∴ 160 × 25/100
= 40
Thus, the value will be 40.
(d) 80% of 240
∴ 240 × 80/100
= 182
Hence, the value will be 182
(e) 5% of 500
= 500 × 5/100
= 25
Thus, the value will be 25
(f) 10% of 25
∴ 25 × 10/100
= 2.5
Hence, the value will be 2.5
Question no – (48)
Solution :
In the question we get,
A man spends 6 1/4% of what he has.
he has spent Rs 62.50
Let, he had x amount of money,
∴ x × 25/100/4 = 62.50
= x = 625/10 × 100 × 4/25
= 1000 Rs.
Hence, he had 1000 Rs.
Question no – (49)
Solution :
In the given question,
Population in 1996 = 51,69,420
Population in 1997 = 54,27,891
First, Increase value,
= 55427891 – 5169420
= 258471
Now in percent,
= 258471/5169420 × 100
= 5%
Question no – (50)
Solution :
According to the question,
In an examination Manish gets = 60% marks.
Total number of marks is = 150
Now, = 150 × 60/100
= 90
So, Manish gets 90 marks.
Question no – (51)
Solution :
In the question we get,
A steel almirah is priced at = Rs 1640.
The price goes up by = 12%
Now,
= 1640 × 12/100
= 196.8
∴ New price of steel almirah,
= 1640 + 196
= 1836.80 rupees
Hence, the new price of steel almirah will be 1836.80 rupees.
Question no – (52)
Solution :
First, He spend in total
= (320 + 5.30 + 10)
= 335.30 rupees
Now, his Profit,
= 350.75 – 335.30
= 15.45 rupees
So, he will earn 15.45 rupees.
Question no – (53)
Solution :
In the question we get,
A man sold a refrigerator for = Rs 4100.00
Lost = Rs 300.00.
Now, Cost price,
4100 + 300
= 4400 Rs.
Thus, the cost price of the refrigerator will be 4400 Rs.
Question no – (54)
Solution :
According to the question,
Car was sold by a car dealer for = Rs 43,000.00,
Making a profit of = Rs 3,000.00
Now, Cost price of the car,
= 43000 – 3000
= 40000
Thus, the cost price of the car will be 40000 Rs.
Question no – (55)
Solution :
In the question we get,
Ramesh bought an old motorcycle for = 2,000 Rs
Spent on its overhauling = Rs. 575
Then he sold it for = Rs 2,900
First, C.P in total,
= 2000 + 575
= 2575
Now Profit,
= 2900 – 2575
= 325
Hence, he will make Rs. 325
Question no – (56)
Solution :
First, Interest,
= 3000 × 6 × 2/100
= 360
Now the amount,
= 3000 + 360
= 3360 Rs.
Thus, she will get Rs. 3360 after 2 years.
Question no – (57)
Solution :
(a) The interest he paid at the end of the first year
Principal = 3000,
rate = 4%
∴ Interest = 3500 × 4 × 1/100
= 120 Rs
(b) The total amount he paid at the end of the 1st year.
= Total amount,
= 100 + 120
= 1120 Rs.
Question no – (58)
Solution :
17th Feb 1996 to 24 Sep 1998
= 10 m 12 day + 1 year + 8 m 2 year days
= 2 years 7 m 6 day
= 949/365 year
= 2.6 year
∴ Interest,
= 2300 × 21/2 × 2,6/1000
= 627.9 Rs.
∴ Amount,
= 2300 + 627.9
= 2927.9 Rs.
Thus, Arvind returned to Puneet 2927.9 Rs.
Question no – (59)
Solution :
2 year 3 month ….(given)
= 27 m
= 27/12 year
Now, the Interest,
= 8000 × 25/2 × 22/12/100
= 2250 Rs.
∴ Amount,
= 8000 + 2250
= 10250 Rs.
Hence, he will receive 10250 Rs.
Question no – (60)
Solution :
According to the question,
Anil deposited Rs 800 at 6% for 3 years.
Rohit deposited Rs 900 at 5% for 3 years
Now, Anil’s Interest,
= 800 × 6 × 3/100
= 144
Rohit Interest,
= 900 × 5 × 3/100
= 135
Therefore, Anil will get more interest.
Question no – (61)
Solution :
In the question,
Amount =365 Rs
Rate =8% for
Time =300 days
∴ Simple interest,
= 265 × 8 × 300/365/100
= 24 Rs.
Hence, the simple interest will be 24 Rs.
Question no – (62)
Solution :
Given in the question,
Ashok borrowed Rs 3,000 from post office at 4%
amount paid by him on end of 2 1/2% years = ?
First, Interest,
= 3000 × 4 × 5/2/100
= 300 Rs.
Now the Amount,
= 3000 + 300
= 3300 Rs.
Thus, he will pay 3300 Rs.
Question no – (63)
Solution :
In the question we get,
Ritika deposited a sum of Rs 1,000
interest paid by the post office is 12%
she will receive at the end of 3 years = ?
First, Interest
= 100 × 12 × 3/100
= 360 Rs.
Now, Amount
= 1000 + 360
= 1360 Rs.
Thus, she will receive Rs. 1360 at the end of 3 years.
Question no – (64)
Solution :
1st January to 15th march, ….(Given)
= 30 + 28 + 15
= 73 days
∴ 73/365 year
Now, Simple Interest,
= 75,50 × 63 × 73/365/100 × 1
= 69/100
= 0.69
∴ The Amount,
= 57.50 + 0.69
= 58.19 Rs.
Question no – (65)
Solution :
From the question we get,
Manish borrowed = Rs 650
At = 8% per annum
returned money after = 6 months
First, the Interest,
= 650 × 8 × 6/12/100
= 26 Rs.
Now the Amount,
= 650 + 26
= 676 Rs.
Thus, he repay to Puneet 676 Rs.
Question no – (66)
Solution :
In the question we get,
Sonam borrows Rs 400 from her friend.
Interest of 5% per annum.
First, the Interest,
= 400 × 2 × 5/100
= 40 Rs.
∴ She will pay (amount)
= 400 + 40
= 440 Rs.
So, she will pay back 440 Rs.
Question no – (67)
Solution :
(a) A diameter is the longest chord of the circle.
= Given statement is True
(b) If we join any two points on a circle, we get a diameter of the circle.
= Given statement is False
(c) A diameter of a circle contains the centre of the circle.
= Given statement is True
(d) The length of a radius of a circle is twice the length of diameter of the circle.
= Given statement is False
(e) A semi-circle is an arc.
= Given statement is True
(f) The length of a circle is called its circumference.
= Given statement is True
Question no – (68)
Solution :
As we know that,
Sum of angles are 180° then it is a supplementary angle.
Now,
(a) 91°. 89°
= 91° + 89°
= 180 [supplementary]
(b) 100° 70°
= 100° + 70°
= 170° [Not supplementary]
(c) 125°, 55°
= 125 + 55
= 180°
= [Supplementary]
(d) 115°, 65°
= 115 + 65
= 180° [Supplementary]
(e) 63°, 27°.
= 63 + 27
= 90°
= [Not supplementary]
Question no – (69)
Solution :
As we know that,
If sum of angle are 90° then it is a complement angle.
Now,
(a) 41°, 59°
= 41 + 59
= 100° [Not complement]
(b) 63°, 27°
= 63 + 27
= 90° (Complement )
(c) 81° 9°
= 81 + 9
= 90° (complement )
(d) 70°, 110°
= 70 + 110°
= 180° (Not complement)
(e) 70°, 20°.
= 70° + 20°
= 90 (Not complement)
Question no – (70)
Solution :
Let, the two equal angle are = 50°
∴ 50 + 50 x = 180°
= x = 80
Therefore, the required angles are, = 80°, 50°, 50°
Question no – (72)
Solution :
(i) Circles whose radii is = 7 cm.
We know, Circumference of circle = 2 π r
∴ 2 π r
= 2 × 22/7 × 7
= 44 cm
Thus, the circumference of the circle will be 44 cm.
(ii) Circle whose radii is = 21 cm.
We know that,
Circumference of circle = 2 π r
∴ Circumference,
= 2 π r = 2 × 22/7 × 21
= 132 cm
Hence, the circumference of the circle will be 132 cm.
Question no – (73)
Solution :
In the given question,
One of the angle of a parallelogram is 65°
Other angle,
= 180° – 65°
= 115°
∴ Other angles are,
= 115°, 115°, 65°,65°
Question no – (75)
Solution :
In the question we get,
Two angles of a triangle are = 70° and 45°
∴ Third angle
= 180° – (70° + 45°)
= 180° – 115°
= 65°
Hence, the third angle will be 65°
Question no – (76)
Solution :
In the question,
One of the two equal angles of a triangle is 70°
∴ Other angle,
= 180° – (70° + 70° )
= 180° – 140°
= 40°
Hence, the other angle will be 40°
Question no – (77)
Solution :
According to the question,
Three angles of a quadrilateral are = 70°, 110° and 90°
∴ The Fourth angle,
= 360° – (110° + 70 + 90°)
= 360° – 270°
= 90°
Thus, the fourth angle will be 90°
Question no – (78)
Solution :
In the question we get,
length of a rectangular field = 60 m
breadth of a rectangular field = 15 m
We know that,
Area of rectangle = length × breadth
∴ Area of rectangle,
= 60 × 15
= 400 m2
∴ Cost of ploughing
= 900 × 1.5/10
= 1350 Rs.
Question no – (79)
Solution :
In the question,
The side of a square field is 82 m
∴ Area of square,
= (side)2
= (82)2
= 6724 m2
Now, Cost of lying grass,
= 6724 × 25/100
= 1681 Rs.
Hence, the cost of laying grass will be 1681 Rs.
Question no – (80)
Solution :
Area of square tile,
= (10)2 = 100 cm2
= 100/100 × 100 m2
Area of wall,
= 2.5 × 2
= 5 m2
∴ Tile required,
= 5/100/100 × 100
= 5 × 100 × 100/100
= 5000 litres
= 500 tiles will be required
∴ Cost of covering the wall,
= 500/12 × 4
= 166.666666667
Question no – (81)
Solution :
Value of cube,
= (4)3 = 64 m3
= 64 × 100 × 100 × 100 cm3
∴ Volume of rectangle,
= 20 × 10 × 5 cm3
∴ Number of block,
= 64 × 100 × 100 × 100/10 × 20 × 5
= 64000 places
Question no – (82)
Solution :
Volume of rectangular lay,
= 8 × 1.5 × 75 × 100 × 100 cm3
∴ Volume of block,
= (25)3
= 25 × 25 × 25 cm3
∴ Number of cubical block,
= 8 × 1.5 × 75 × 100 × 100/25 × 25 × 25 × 10
= 576 blocks
Question no – (87)
Solution :
Volume of blocks,
= 3.5/10 × 30/100 × 15/100 m3
= 35 × 3 × 15/100 × 100 m3
Now the Cost block of wood,
= 1800 × 35 × 3 ×15/100 × 100
= 567/2
= 283.50 Rs.
Hence, the cost of block of wood will be 283.50 Rs.
Question no – (88)
Solution :
Volume of cube,
= (20)3
= 8000 cm3
Another wood,
= 10 × 20 × 15
= 3000 cm3
∴ Cubical piece will more heavier.
Question no – (89)
Solution :
In the question we get,
The volume of a rectangular solid is = 225 cu.cm.
Its height is 5 cm and breadth is = 10 cm
Let, its length = x cm
∴ x × 5 × 10 = 225
= x = 45/10
= x = 4.5 cm
Therefore, the length of rectangular solid will be 4.5 cm.
Question no – (90)
Solution :
In the given question we get,
length is 18 cm,
breadth 10 cm
height 6 cm.
∴ Volume of soil brick,
= 18 × 10 × 6
= 1080 cm3
Thus, the volume of a solid brick will be 1080 cm3
Question no – (91)
Solution :
Volume of each bricks,
= 20 × 10 × 5 cm3
∴ Volume of each wall brick,
= 10 × 100 × 4 × 100 × 50 cm3
∴ Bricks required,
= 100 × 100 × 4 × 100 50/20 × 10 × 5
= 20000 brick
Hence, 20000 brick will be required.
Question no – (92)
Solution :
In the question we get,
Cube whose edge is 2.5 m.
∴ Volume of cube,
= (edge)3
= (2.5)3
= 15.625 m3
Hence, the volume of a cube will be 15.625 m3
Question no – (93)
Solution :
First, Volume of soap cake,
= 7 × 5 × 2.5 cm3
Now, Volume of card-board box,
= 56 × 49 × 25 cm3
Therefore, Soap cakes can be placed,
= 56 × 49 × 25 × 10/7 × 5 × 2.5
= 784 cakes
Question no – (95)
Solution :
(a) The amount of space a solid occupies is called Volume.
(b) The volume of a solid pipe and a hollow pipe of the same shape and size are Equal.
(c) If a rectangular box is filled in by fourteen 1 cm cubes, its volume is 14 cm3.
(d) The volume of a 1 cm cube is 1 cm3.
(e) The volume of a rectangular box (or a cuboid) = Length × breadth × height
Question no – (96)
Solution :
Volume of brick,
= 5 cm × 4 cm × 2 cm …(according to the question)
= 40 cm3
∴ Volume of wall,
= 10 × 5 × 20 × 100 × 100 cm3
∴ Brick required,
= 10 × 5 × 20 × 100 × 100/40
= 250000 bricks
Hence, 250000 bricks will be required.
Question no – (97)
Solution :
In the given question,
Length is = 0.01 m,
Breadth is = 6 cm
Height is = 5 cm
∴ Volume of ice-cream brick,
= 1.01/100 × 100 × 6 × 5 cm
= 30 cm3
Hence, the volume of ice-cream brick will be 30 cm3
Question no – (98)
Solution :
Given in the question,
Cuboid with dimensions = 15 m × 10 m × 8 m
∴ The Volume,
= 15 × 10 × 8
= 1200 cm3
So, the volume of the space occupies by it will be 1200 cm3
Question no – (100)
Solution :
(i) The biggest side of an obtuse-angles triangle is opposite the Obtuse angle.
(ii) In a scalene triangle, all the three sides are unequal.
(iii) In an equilateral triangle, all the three sides are equal.
(iv) The sum of any two Side of a triangle is greater then third angle.
(v) In an right-angled triangle, the other two angles of the triangle must be Acute angle.
Question no – (101)
Solution :
Given in the question,
The measures of the non-equal angle of an isosceles triangle is 40°
Let, than two angles are x
∴ x + x + 40 = 180°
= 2x = 140°
= x = 70°
∴ Angle are = 70°, 70°, 40°
Next Chapter Solution :
👉 Chapter 2 👈
